R E S E A R C H
Open Access
A new iterative algorithm for the sum of two
different types of finitely many accretive
operators in Banach space and its connection
with capillarity equation
Li Wei
*and Liling Duan
*Correspondence:
[email protected] School of Mathematics and Statistics, Hebei University of Economics and Business, Shijiazhuang, 050061, China
Abstract
In this paper, we present a new iterative algorithm with errors to solve the problems of finding zeros of the sum of finitely manym-accretive operators and finitely many
α
-inversely strongly accretive operators in a real smooth and uniformly convex Banach space. Strong convergence theorems are established, which extend the corresponding works given by some authors. Moreover, the relationship among the zero of the sum ofm-accretive operator andα
-inversely strongly accretive operator, the solution of one kind variational inequality, and the solution of the capillarity equation is investigated.MSC: 47H05; 47H09; 47H10
Keywords:
α
-inversely strongly accretive operator; sum; zero; iterative algorithm; strong convergence; variational inequality; capillarity equation1 Introduction and preliminaries
LetEbe a real Banach space with norm · and letE∗denote the dual space ofE. We use ‘→’ and ‘’ to denote strong and weak convergence either inEor inE∗, respectively. We denote the value off ∈E∗atx∈Ebyx,f.
A Banach spaceEis said to be uniformly convex if, for eachε∈(, ], there existsδ> such that
x=y= , x–y ≥ε ⇒ x+y
≤ –δ.
A Banach spaceEis said to be smooth if
lim t→
x+ty–x t
exists for eachx,y∈ {z∈E:z= }.
In addition, we define a functionρE: [, +∞)→[, +∞) called the modulus of
smooth-ness ofEas follows:
ρE(t) =sup
x+y+x–y– :x,y∈E,x= ,y ≤t
.
It is well known thatEis uniformly smooth if and only if ρE(tt) →, ast→. Letq> be a real number. A Banach spaceE is said to beq-uniformly smooth if there exists a positive constantCsuch thatρE(t)≤Ctq. It is obvious thatq-uniformly smooth Banach
space must be uniformly smooth.
The normalized duality mappingJ:E→E∗is defined by
Jx:=f∈E∗:x,f=x=f, x∈E.
It is well known thatJis single-valued and norm-to-norm uniformly continuous on each bounded subsets ofEifEis a real smooth and uniformly convex Banach space; see []. Moreover,J(cx) =cJx, for allx∈Eandc∈R. In what follows, we still denote byJthe single-valued normalized duality mapping. If,Eis reduced to the Hilbert spaceH, then
J≡Iis the identity mapping. The normalized duality mappingJis said to be weakly se-quentially continuous if{xn}is a sequence inEwhich converges weakly toxit follows that
{Jxn}converges in weak∗toJx.Jis said to be weakly sequentially continuous at zero if{xn}
is a sequence inEwhich converges weakly to it follows that{Jxn}converges in weak∗
to .
LetCbe a nonempty, closed, and convex subset ofEand letQbe a mapping ofEontoC. ThenQis said to be sunny [] ifQ(Q(x) +t(x–Q(x))) =Q(x), for allx∈Eandt≥.
A mappingQofEinto Eis said to be a retraction [] ifQ=Q. If a mappingQis a retraction, thenQ(z) =zfor everyz∈R(Q), whereR(Q) is the range ofQ.
For a mappingU:C→C, we useFix(U) to denote the fixed point set of it; that is,
Fix(U) :={x∈C:Ux=x}.
For an operatorA:D(A)⊂E→E, we useA– to denote the set of zeros of it; that is,
A– :={x∈D(A) :Ax= }.
LetT:C→Ebe a mapping. ThenTis said to be () nonexpansive if
Tx–Ty ≤ x–y, for∀x,y∈C;
() k-Lipschitz if there existsk> such that
Tx–Ty ≤kx–y, for∀x,y∈C.
In particular, if <k< , thenTis called a contraction and ifk= , thenTreduces to a nonexpansive mapping;
() accretive if for allx,y∈C, there existsj(x–y)∈J(x–y)such that
Tx–Ty,j(x–y)≥,
whereJis the normalized duality mapping;
() α-inversely strongly accretive if for allx,y∈C, there existsj(x–y)∈J(x–y)such that
() m-accretive ifT is accretive andR(I+λT) =E, for∀λ> ;
() strongly positive (see []) ifEis a real smooth Banach space and there existsγ> such that
Tx,Jx ≥γx, for∀x∈C.
In this case,
aI–bT= sup
x≤
(aI–bT)x,J(x),
whereIis the identity mapping anda∈[, ],b∈[–, ]. We denote byJA
r (forr> ) the resolvent of the accretive operatorA; that is,JrA:= (I+
rA)–. It is well known thatJA
r is nonexpansive andFix(JrA) =A–.
A subsetCofEis said to be a sunny nonexpansive retract ofEif there exists a sunny nonexpansive retraction ofEontoC.
Many practical problems can be reduced to finding zeros of the sum of two accretive op-erators; that is, ∈(A+B)x. Forward-backward splitting algorithms, which have recently received much attention to many mathematicians, were proposed by Lions and Mercier [], by Passty [], and, in a dual form for convex programming, by Han and Lou [].
The classical forward-backward splitting algorithm is given in the following way:
xn+= (I+rnB)–(I–rnA)xn, n≥. ()
Based on iterative algorithm (), much work has been done for findingx∈Hsuch that
x∈(A+B)–, whereAandBareα-inversely strongly accretive operator andm-accretive operator defined in the Hilbert spaceH, respectively. However, most of the existing work are undertaken in the frame of Hilbert spaces; see [–],etc.
Recently, Qinet al., presented the following iterative algorithm in the frame ofq -uni-formly smooth Banach spacesEin []:
x∈C, xn+=αnf(xn) +βn(I+rnB)–
(I–rnA)xn+en
+γnfn, n≥, ()
where{en}is the error sequence,f is a contraction,AandBareα-inversely strongly
accre-tive operator andm-accretive operator, respectively. If (A+B)–=∅, they proved that{x
n}
converges strongly tox=Q(A+B)–f(x), whereQ(A+B)–is the unique sunny nonexpansive
retraction ofEonto (A+B)–, under some conditions.
On the other hand, there are some excellent work done on approximating fixed points of nonexpansive mappings. For example, in , Yaoet al.presented the following iterative algorithm in the frame of Hilbert space in []:
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
x∈C,
yn=PC[( –αn)xn],
xn+= ( –βn)xn+βnTyn, n≥,
()
wherePCis the metric projection fromHontoCandT:C→Cis a nonexpansive
map-ping withFix(T)=∅. They proved that{xn}constructed by () converges strongly to a
In , Marino and Xu, presented the following iterative algorithm in the frame of Hilbert spaces in []:
x∈C, xn+=αnγf(xn) + (I–αnA)Txn, n≥, ()
wheref is a contraction,Ais a strongly positive linear bounded operator, andTis nonex-pansive. IfFix(T)=∅, they proved that{xn}converges strongly top∈Fix(T), which solves
the variational inequality(γf–A)p,z–p ≤, for∀z∈Fix(T), under some conditions. Our paper is organized in the following way: in Section , inspired by the work in [– ], we shall present the following iterative algorithm with errors in a real smooth and uniformly convex Banach space:
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
x∈C,
yn=QC[( –αn)(xn+en)],
zn= ( –βn)xn+βn[ayn+
N
i=aiJrAni,i(yn–rn,iBiyn)],
xn+=γnηf(xn) + (I–γnT)zn, n≥,
(A)
whereC is a nonempty, closed, and convex sunny nonexpansive retract ofE,QC is the
sunny nonexpansive retraction ofEontoC,{en} ⊂Eis the error sequence,{Ai}Ni= is a fi-nite family ofm-accretive operators and{Bi}Ni= is a finite family ofα-inversely strongly accretive operators.T :E→Eis a strongly positive linear bounded operator with coef-ficientγ andf :E→Eis a contraction with coefficientk∈(, ).JAi
rn,i= (I+rn,iAi)–, for
i= , , . . . ,N,Nm=am= , <am< , form= , , , . . . ,N. More detail of iterative
al-gorithm (A) will be presented in Section . Then{xn}is proved to converge strongly to
p∈
N
i=(Ai+Bi)–, which is also a solution of one kind variational inequality.
Our main contributions in Section are:
(i) the discussion is undertaken in the frame of real smooth and uniformly convex Banach space, which is more general than that in Hilbert space or inq-uniformly smooth Banach space;
(ii) the assumption that ‘the normalized duality mappingJis weakly sequentially continuous’ in most of the existing related work is weaken to ‘Jis weakly sequentially continuous at zero’;
(iii) a new path convergence theorem (Lemma ) is obtained which is a direct extension of the corresponding result in [] from Hilbert space to real smooth and uniformly convex Banach space;
(iv) the connection between zeros of the sum ofm-accretive operators andα-inversely strongly accretive operators and the solution of one kind variational inequalities is being set up.
In Section , one kind capillarity equation is discussed, from which we can see the con-nection among the unique solution of this equation, the unique solution of one kind vari-ational inequality and the iterative algorithm presented in Section .
Next, we list some results we need in sequel:
Lemma (see []) Let E be a real uniformly convex Banach space,C be a nonempty,
closed, and convex subset of E and T :C →E be a nonexpansive mapping such that
Fix(T)=∅,then I–T is demiclosed at zero.
Lemma (see []) In a real Banach space E,the following inequality holds:
x+y≤ x+ y,j(x+y), ∀x,y∈E,
where j(x+y)∈J(x+y).
Lemma (see []) Let{an}and{cn}be two sequences of nonnegative real numbers
satis-fying
an+≤( –tn)an+bn+cn, ∀n≥,
where {tn} ⊂ (, ) and {bn} is a number sequence. Assume that
∞
n=tn = +∞, lim supn→∞bn
tn ≤,and
∞
n=cn< +∞.Thenlimn→∞an= .
Lemma (see []) Let E be a Banach space and let A be an m-accretive operator.For
λ> ,μ> ,and x∈E,one has
Jλx=Jμ
μ λx+
–μ λ
Jλx
,
where Jλ= (I+λA)–and Jμ= (I+μA)–.
Lemma (see []) Let E be a real Banach space and let C be a nonempty,closed,and convex subset of E. Suppose A:C →E is a single-valued operator and B:E→E is
m-accretive.Then
Fix(I+rB)–(I–rA)= (A+B)–, for∀r> .
Lemma (see []) Assume T is a strongly positive bounded operator with coefficientγ >
on a real smooth Banach space E and <ρ≤ T–.ThenI–ρT ≤ –ργ. 2 Strong convergence theorems
Lemma Let E be a real smooth and uniformly convex Banach space and C be a nonempty,closed,and convex sunny nonexpansive retract of E,and let QC be the sunny
nonexpansive retraction of E onto C.Let f :E→E be a fixed contractive mapping with coefficient k∈(, ),T:E→E be a strongly positive linear bounded operator with coef-ficientγ and U:C→C be a nonexpansive mapping.Suppose that the duality mapping J:E→E∗is weakly sequentially continuous at zero, <η<γk andFix(U)=∅.If for each t∈(, ),define Tt:E→E by
Ttx:=tηf(x) + (I–tT)UQCx, ()
then Tthas a fixed point xt,for each <t≤ T–,which is convergent strongly to the fixed
variational inequality:for∀z∈Fix(U),
(T–ηf)p,J(p–z)
≤. ()
Proof Step .Ttis a contraction, for <t<T–.
In fact, noticing Lemma , we have
Ttx–Tty ≤tηf(x) –f(y)+(I–tT)(UQCx–UQCy)
≤ktηx–y+ ( –tγ)x–y
= –t(γ –kη)x–y,
which implies thatTtis a contraction since <η<γk.
Then Lemma implies thatTt has a unique fixed point, denoted byxt, which uniquely
solves the fixed point equationxt=tηf(xt) + (I–tT)UQCxt.
Step .{xt}is bounded, fort∈(,T–).
Forp∈Fix(U)⊂C, we havep=UQCp, then
xt–p=(I–tT)(UQCxt–p) +t
ηf(xt) –Tp
≤( –tγ)xt–p+tηf(xt) –Tp
= ( –tγ)xt–p+tη
f(xt) –f(p)
+ηf(p) –Tp
≤( –tγ)xt–p+t
kηxt–p+ηf(p) –Tp
= –t(γ –kη)xt–p+tηf(p) –Tp.
This ensures that
xt–p ≤
ηf(p) –Tp
γ –kη .
Thus{xt}is bounded, which implies that both{f(xt)}and{TUQCxt}are bounded.
Step .xt–UQCxt→, ast→.
Noticing the result of Step , we havext–UQCxt=tηf(xt) –TUQCxt →, ast→.
Step .(T–ηf)x– (T–ηf)y,J(x–y) ≥(γ–kη)x–y, for∀x,y∈E. In fact,
(T–ηf)x– (T–ηf)y,J(x–y)
= Tx–Ty,J(x–y)–ηf(x) –f(y),J(x–y)
≥γx–y–kηx–y= (γ–kη)x–y.
Step . If the variational inequality () has a solution, then the solution must be unique. Suppose bothu∈Fix(U) andv∈Fix(U) are the solutions of the variational inequality (). Then we have
(T–ηf)v,J(v–u)
and
(T–ηf)u,J(u–v)
≤. ()
Adding up () and (), we obtain
(T–ηf)u– (T–ηf)v,J(u–v)
≤.
In view of the result of Step , we haveu=v.
Step .xt→p∈Fix(U), ast→, which satisfies the variational inequality (). For∀z∈Fix(U),xt–z=t(ηf(xt) –Tz) + (I–tT)(UQCxt–z). Thus Lemma implies that
xt–z≤ I–tTUQCxt–UQCz+ tηf(xt) –Tz,J(xt–z)
≤( –tγ)xt–z+ tηf(xt) –Tz,J(xt–z)
.
Then
xt–z≤
γ ηf(xt) –Tz,J(xt–z)
= γ
ηf(xt) –f(z),J(xt–z)
+ ηf(z) –T(z),J(xt–z)
≤
γ
ηkxt–z+ ηf(z) –Tz,J(xt–z)
.
Therefore, for∀z∈Fix(U), we have
xt–z≤
γ – kηηf(z) –Tz,J(xt–z)
. ()
Since{xt} is bounded as t→+, we can choose{tn} ⊂(, ) such that tn→+ and
xtnp. From Lemma and the result of Step , we see thatp=UQCp=Up. Thus
p∈Fix(U). Substitutingzbypin (), then we can deduce thatxtn→psinceJis weakly sequentially continuous at zero. Next, we shall prove thatpsolves the variational inequal-ity ().
Sincext=tηf(xt) + (I–tT)UQCxt,
(T–ηf)xt= –
t(I–tT)(I–UQC)xt.
For∀z∈Fix(U), sinceUis nonexpansive, (T–ηf)xt,J(xt–z)
= –
t (I–tT)(I–UQC)xt,J(xt–z)
= –
t (I–UQC)xt– (I–UQC)z,J(xt–z)
+ T(I–UQC)xt,J(xt–z)
= –
t
xt–z– UQCxt–UQCz,J(xt–z)
+ T(I–UQC)xt,J(xt–z)
≤ T(I–UQC)xt,J(xt–z)
Sincextn→p, we have (I–UQC)xtn →(I–UQC)p= , as n→ ∞. Since{xtn} is bounded, (T–ηf)xtn→(T–ηf)pandJis uniformly continuous on each bounded sub-set of E, taking the limits on both sides of () we have(T–ηf)p,J(p–z) ≤, for
z∈Fix(U). Thuspsatisfies ().
In a summary, we infer that each cluster point of{xt}is equal top, which is the unique
solution of the variational inequality ().
This completes the proof.
Remark Lemma is a direct extension of Theorem . in [] from Hilbert space to real smooth and uniformly convex Banach space.
Theorem Let E be a real smooth and uniformly convex Banach space and C be a nonempty,closed,and convex sunny nonexpansive retract of E,and let QC be the sunny
nonexpansive retraction of E onto C.Let f :E→E be a fixed contractive mapping with coefficient k∈(, ),T:E→E be a strongly positive linear bounded operator with coeffi-cientγ.Suppose that the duality mapping J:E→E∗is weakly sequentially continuous at zero,and <η<γk.Let Ai:C→Ebe m-accretive operator and Bi:C→E beα-inversely
strongly accretive operator,where i= , , . . . ,N.Suppose that,for∀r> and i= , , . . . ,N,
Bix–Biy,J
(I–rBi)x– (I–rBi)y
≥.
Let {xn} be generated by the iterative algorithm (A), <am < ,for m= , , , . . . ,N,
N
m=am= .Suppose{en}∞n=⊂E,{αn}, {βn},and{γn}are three sequences in(, )and
{rn,i} ⊂(, +∞)satisfying the following conditions:
(i) ∞n=γn=∞,γn→, γγn–n →,βn→,αn→,asn→ ∞; (ii) ∞n=|αn+–αn|< +∞,
∞
n=|βn+–βn|< +∞,
∞
n=( –γnγ)αnβn< +∞;
(iii) ∞n=|rn+,i–rn,i|< +∞andrn,i≥ε> ,forn≥andi= , , . . . ,N;
(iv) ∞n=en< +∞.
IfNi=(Ai+Bi)–=∅,then{xn}converges strongly to a point p∈
N
i=(Ai+Bi)–,which
is the unique solution of the following variational inequality:for∀z∈Ni=(Ai+Bi)–,
(T–ηf)p,J(p–z)
≤. (∗)
Proof Letun,i= (I–rn,iBi)yn,vn=ayn+
N
i=aiJrAni,iun,i, forn≥, wherei= , , . . . ,N. We shall split the proof into five steps:
Step .{xn},{un,i}(i= , , . . . ,N),{yn},{vn}, and{zn}are all bounded.
From the assumptions onBi, in view of Lemma , we have, for∀x,y∈C,
(I–rBi)x– (I–rBi)y
≤ x–y– rBix–Biy,J
(I–rBi)x– (I–rBi)y
≤ x–y,
which implies that (I–rBi) is nonexpansive, forr> .
Then noticing the facts that both (I–rn,iBi) andJrAni,iare nonexpansive, fori= , , . . . ,N, we have, for∀p∈Ni=(Ai+Bi)–⊂C,
Using Lemma , we have, forp∈Ni=(Ai+Bi)–,
zn–p ≤( –βn)xn–p+βn
ayn–p+ N
i=
aiJrAni,i(I–rn,iBi)(yn–p)
≤( –βn)xn–p+βnayn–p+βn N
i=
aiyn–p
= ( –βn)xn–p+βnyn–p. ()
Then Lemma implies that forn≥,
xn+–p ≤γnηf(xn) –Tp+(I–γnT)(zn–p)
≤γnηkxn–p+γnηf(p) –Tp+ ( –γnγ)zn–p. ()
Noticing ()-(), we have, forn≥,
xn+–p
≤γnηkxn–p+γnηf(p) –Tp
+ ( –γnγ)
( –βn)xn–p+βnyn–p
≤γnηk+ ( –γnγ)( –βn) + ( –γnγ)( –αn)βn
xn–p+γnηf(p) –Tp
+ ( –γnγ)βnen+ ( –γnγ)βnαnen+p
= –αnβn( –γnγ) –γn(γ –kη)
xn–p+γnηf(p) –Tp
+ ( –γnγ)βnen+ ( –γnγ)αnβnen+p
≤ –γn(γ –kη)
xn–p+γnηf(p) –Tp+ en+ ( –γnγ)αnβnp
≤max
xn–p,
ηf(p) –Tp
γ–kη
+ en+ ( –γnγ)αnβnp. ()
By using the inductive method, we can easily get the following result from ():
xn+–p ≤max
x–p,
ηf(p) –Tp
γ–kη
+
n
k=
ek+p n
k=
( –γkγ)αkβk.
Therefore, from assumptions (ii) and (iv), we know that{xn}is bounded.
For∀p∈Ni=(Ai+Bi)–, sinceyn–p ≤ ( –αn)(xn+en) –p ≤ xn+en+p,
{yn}is bounded, which implies that{un,i}is bounded in view of the fact thatI–rn,iBiis
nonexpansive, for eachi= , , . . . ,N. Moreover,{JAi
rn,i(I–rn,iBi)yn}is bounded sinceJ
Ai
rn,iis nonexpansive, fori= , , . . . ,N. Thus
{vn}is bounded, which ensures that{zn} is bounded. Sincern,i≥ε> ,Biyn=yn–rnu,in,i is
bounded, forn≥ andi= , , . . . ,N.
Set M =sup{un,i,JrAni,iun,i,Biyn,xn,vn,zn,Tzn,ηf(xn): n≥,i = , , . . . ,N}.
First, we shall discussJAi
rn,iun,i–J
Ai
rn–,iun–,i, forn≥. Ifrn–,i≤rn,i, then by using Lemma , we have
JAi
rn,iun,i–J
Ai
rn–,iun–,i
=JAi
rn–,i
rn–,i
rn,i
un,i+
–rn–,i
rn,i
JAi
rn,iun,i
–JAi
rn–,iun–,i
≤rn–,i
rn,i
un,i+
–rn–,i
rn,i
JAi
rn,iun,i–un–,i
≤rn–,i
rn,i
un,i–un–,i+
–rn–,i
rn,i
JAi
rn,iun,i–un–,i
≤ un,i–un–,i+
rn,i–rn–,i
ε J
Ai
rn,iun,i–un–,i. ()
Ifrn,i≤rn–,i, then imitating the proof of (), we have
JAi
rn,iun,i–J
Ai
rn–,iun–,i≤ un,i–un–,i+
rn–,i–rn,i
ε J
Ai
rn,iun,i–un–,i. ()
Combining () and (), we have, forn≥,
JAi
rn,iun,i–J
Ai
rn–,iun–,i
≤ un,i–un–,i+|
rn–,i–rn,i|
ε J
Ai
rn,iun,i–un–,i
≤ un,i–un–,i+
|rn–,i–rn,i|
ε M
≤(I–rn,iBi)(yn–yn–)+|rn,i–rn–,i|Biyn–+
|rn–,i–rn,i|
ε M
≤ yn–yn–+|rn,i–rn–,i|Biyn–+
|rn–,i–rn,i|
ε M. ()
LetM= (ε+ )M, and using (), we have, forn≥,
vn–vn– ≤ayn–yn–+
N
i=
aiJrAni,iun,i–J
Ai
rn–,iun–,i
≤ yn–yn–+M
N
i=
ai|rn,i–rn–,i|. ()
Using (), we have, forn≥,
zn–zn– ≤( –βn)xn–xn–+|βn–βn–|xn– +βnvn–vn–+|βn–βn–|vn–
≤( –βn)xn–xn–+|βn–βn–|xn–+βnyn–yn–
+βnM
N
i=
Noticing that forn≥,
yn–yn– ≤( –αn)xn–xn–+|αn–αn–|xn–
+ ( –αn)en–en–+|αn–αn–|en–. () Using () and (), we have, forn≥,
xn+–xn
≤γnηf(xn) –f(xn–)+η|γn–γn–|f(xn–)+I–γnTzn–zn– +|γn–γn–|Tzn–
≤γnηkxn–xn–+η|γn–γn–|f(xn–)+ ( –γnγ)zn–zn– +|γn–γn–|Tzn–
≤( –γnγ)( –αnβn) +γnηk
xn–xn–+ M|γn–γn–|
+ M( –γnγ)|βn–βn–|+ ( –γnγ)Mβn N
i=
ai|rn,i–rn–,i|
+ ( –γnγ)βn|αn–αn–|
M+en–
+ ( –γnγ)βn( –αn)en–en–
≤ –γn(γ –ηk)
xn–xn–+ M|γn–γn–|
+ M( –γnγ)|βn–βn–|+ ( –γnγ)Mβn N
i=
ai|rn,i–rn–,i|
+ ( –γnγ)βn|αn–αn–|
M+en–
+ ( –γnγ)βn( –αn)en–en–. () From the assumptions on{αn},{βn},{γn},{rn,i}, and{en}, in view of (), and Lemma ,
we havelimn→∞xn+–xn= .
Step . SetWn= [aI+
N
i=aiJrAni,i(I–rn,iBi)], thenWn:C→Cis nonexpansive and Fix(Wn) =
N
i=(Ai+Bi)–.
It is obvious thatWnis nonexpansive and
N
i=(Ai+Bi)–⊂Fix(Wn). So we are left to
show thatFix(Wn)⊂
N
i=(Ai+Bi)–.
In fact, ifp∈Fix(Wn), thenWnp=p. For∀q∈
N
i=(Ai+Bi)–⊂Fix(Wn), we have
p–q ≤ap–q+aJrAn,(I–rn,B)p–q+· · ·+aNJ AN
rn,N(I–rn,NBN)p–q
≤(a+a+· · ·+aN–)p–q+aNJrAnN,N(I–rn,NBN)p–q = ( –aN)p–q+aNJrAnN,N(I–rn,NBN)p–q
≤ p–q.
Therefore,p–q= ( –aN)p–q+aNJrAnN,N(I–rn,NBN)p–q, which implies thatp–
q=JAN
rn,N(I–rn,NBN)p–q. Similarly,p–q=J
A
rn,(I–rn,B)p–q=· · ·=J AN– rn,N–(I–
rn,N–BN–)p–q. Then p –q = Na
i=ai(J
A
rn,(I –rn,B)p–q) + a N
i=ai(J
A
rn,(I –rn,B)p– q) +· · · + aN
N i=ai(J
AN
rn,N(I–rn,NBN)p–q), which implies from the strictly convexity ofEthatp–q=
JrAn,(I–rn,B)p–q=J A
rn,(I–rn,B)p–q=· · ·=J AN
Therefore, JAi
rn,i(I –rn,iBi)p=p, for i = , , . . . ,N. Then p ∈
N
i=(Ai +Bi)–. Thus Fix(Wn)⊂
N
i=(Ai+Bi)–.
Step .Wnyn–yn→, asn→ ∞, whereWnis the same as that in Step .
In fact, since both{xn}and{Wnyn}are bounded andβn→, asn→+∞,
zn–Wnyn= ( –βn)(xn–Wnyn)→, asn→+∞.
Since both{f(xn)}and{Tzn}are bounded andγn→, asn→+∞,
xn+–zn=γn
ηf(xn) –Tzn
→, asn→+∞.
Therefore
xn–Wnyn= (xn–xn+) + (xn+–zn) + (zn–Wnyn)→,
asn→+∞, in view of the fact of Step . Since∞n=en< +∞andαn→, asn→ ∞,
Wnyn–yn=QCWnyn–QC
( –αn)(xn+en)
≤ Wnyn–xn+αnxn+ ( –αn)en →, asn→ ∞.
Moreover,xn+–yn→, asn→ ∞.
Step .lim supn→+∞ηf(p) –Tp,J(xn+–p) ≤, wherep∈
N
i=(Ai+Bi)–, which
is the unique solution of the variational inequality (∗).
Noticing the result of Step and using Lemma , we know that there existsztsuch that
zt=tηf(zt) + (I–tT)WnQCztfort∈(, ). Moreover,zt→p∈Fix(Wn) =
N
i=(Ai+Bi)–,
ast→. And,pis the unique solution of the variational inequality (∗).
Sincezt ≤ zt–p+p, then{zt}is bounded, ast→. Using Lemma , we have
zt–yn =zt–Wnyn+Wnyn–yn
≤ zt–Wnyn+ Wnyn–yn,J(zt–yn)
=tηf(zt) + (I–tT)WnQCzt–Wnyn
+ Wnyn–yn,J(zt–yn)
≤ WnQCzt–Wnyn+ tηf(zt) –TWnQCzt,J(zt–Wnyn)
+ Wnyn–yn,J(zt–yn)
≤ zt–yn+ tηf(zt) –TWnQCzt,J(zt–Wnyn)
+ Wnyn–ynzt–yn,
which implies that
tTWnQCzt–ηf(zt),J(zt–Wnyn)
≤ Wnyn–ynzt–yn.
Sincezt→p,WnQCzt→WnQCp=p, ast→ in view of Step . Noticing the fact that
Tp–ηf(p),J(p–Wnyn)
= Tp–ηf(p),J(p–Wnyn) –J(zt–Wnyn)
+ Tp–ηf(p),J(zt–Wnyn)
= Tp–ηf(p),J(p–Wnyn) –J(zt–Wnyn)
+ Tp–ηf(p) –TWnQCzt+ηf(zt),J(zt–Wnyn)
+ TWnQCzt–ηf(zt),J(zt–Wnyn)
,
we havelim supn→+∞Tp–ηf(p),J(p–Wnyn) ≤.
SinceTp–ηf(p),J(p–xn+)=Tp–ηf(p),J(p–xn+) –J(p–Wnyn)+Tp– ηf(p),J(p–Wnyn)andxn+–Wnyn→, thenlim supn→∞ηf(p) –Tp,J(xn+–p) ≤.
Step .xn→p, asn→+∞, wherep∈
N
i=(Ai+Bi)–⊂Cis the same as that in
Step .
LetM=sup{( –αn)(xn+en) –p:n≥}. By using Lemma again, we have
yn–p≤( –αn)xn–p+ ( –αn)en–αnp,J
( –αn)(xn+en) –p
. ()
Using () and the result of Step , we have
zn–p≤( –βn)xn–p+βnWnyn–Wnp
≤( –βn)xn–p+βnyn–p
≤( –αnβn)xn–p + βn( –αn)en–αnp,J
( –αn)(xn+en) –p
. ()
Using () and Lemma , we have, forn≥,
xn+–p =γn
ηf(xn) –Tp
+ (I–γnT)(zn–p)
≤( –γnγ)zn–p+ γnηf(xn) –Tp,J(xn+–p)
≤( –γnγ)( –αnβn)xn–p+ γnηf(xn) –f(p),J(xn+–p) –J(xn–p)
+ γnηf(xn) –f(p),J(xn–p)
+ γnηf(p) –Tp,J(xn+–p)
+ ( –γnγ)βn( –αn)en,J
( –αn)(xn+en) –p
– αnβn( –γnγ)p,J
( –αn)(xn+en) –p
≤ –γn(γ – ηk)
xn–p+ M
en+αnβn( –γnγ)p
+γn
ηf(p) –Tp,J(xn+–p)
+ ηxn–pxn+–xn
. ()
Letδn()=γn(γ – ηk),δn()=γn[ηf(p) –Tp,J(xn+–p)+ ηxn–pxn+–xn],
Using the assumptions (ii) and (iv), the results of Steps , , and and by using Lemma , we know thatxn→p, asn→+∞.
This completes the proof.
Remark The assumption that ‘theα-inversely strongly accretive operatorBi:C→E
satisfies for∀r> andi= , , . . . ,N,Bix–Biy,J[(I–rBi)x– (I–rBi)y] ≥’ is valid, and
we can find an example in Section (Remark ).
Lemma (see []) Let E be a real q-uniformly smooth Banach space with constant Kq
and C be a nonempty,closed,and convex subset of E.Let A:C→E be anα-inversely strongly accretive operator.Then for∀r≤(qKα
q)
q–, (I–rA)is nonexpansive.
Corollary Let E be a real q-uniformly smooth Banach space with constant Kqand also
be a uniformly convex Banach space.Let C,QC,f,k,η,T,J,Ai,am(m= , , , . . . ,N),γ,
{en},{αn},{βn},{γn},and{rn,i}satisfy the same conditions as those in Theorem.Let Bi:
C→E beα-inversely strongly accretive operator,where i= , , . . . ,N.Let{xn}be generated
by the iterative algorithm(A).Suppose further that
(v) rn,i≤(qKαq)
q–,forn≥andi= , , . . . ,N.
IfNi=(Ai+Bi)–=∅,then{xn}converges strongly to a point p∈
N
i=(Ai+Bi)–,which
is the unique solution of the variational inequality(∗).
Proof Lemma ensures that (I–rn,iBi) is nonexpansive, forn≥ andi= , , . . . ,N. Then
copy the proof of Theorem , the result follows.
This completes the proof.
Corollary If i≡,then iterative algorithm(A)becomes the following one:
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
x∈E,
yn=QC[( –αn)(xn+en)], n≥,
zn= ( –βn)xn+βn[ayn+ ( –a)JrAn(yn–rnByn)], n≥,
xn+=γnηf(xn) + (I–γnT)zn, n≥.
(B)
Let E,C,QC, f,k,η,T,J, γ,{en}, {αn}, {βn},and{γn} satisfy the same conditions as
those in Theorem.Let A:C→E be m-accretive operator and B:C→E beα-inversely strongly accretive operator satisfying that
Bx–By,J(I–rB)x– (I–rB)y≥, for∀r> ,∀x,y∈C.
Suppose that <a< ,{rn} ⊂(, +∞)such that
∞
n=|rn+–rn|< +∞and rn≥ε> for
n≥.
If(A+B)–=∅,then{x
n}generated by the iterative algorithm(B)converges strongly
to p∈(A+B)–,which is the unique solution of the following variational inequality:for
∀z∈(A+B)–,
(T–ηf)p,J(p–z)
Corollary If Bi≡,then iterative algorithm(A)becomes the following one for
approx-imating common zeros of finitely many m-accretive operators:
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
x∈E,
yn=QC[( –αn)(xn+en)], n≥,
zn= ( –βn)xn+βn(ayn+
N
i=aiJrAni,iyn), n≥,
xn+=γnηf(xn) + (I–γnT)zn, n≥.
(C)
Let E,C,QC,f,k,η,T,J,γ,am(m= , , . . . ,N),{en},{αn},{βn},and{γn},{rn,i}satisfy
the same conditions as those in Theorem.Let Ai:C→E be m-accretive operator,i=
, , . . . ,N.
IfNi=A–i =∅,then{xn}generated by(C)converges strongly to a point p∈
N
i=A–i ,
which is the unique solution of the following variational inequality:for∀z∈Ni=A–
i ,
(T–ηf)p,J(p–z)
≤. (∗∗∗)
Corollary If Ai≡,then iterative algorithm(A)becomes to the following one for
ap-proximating common zeros of finitely manyα-inversely strongly accretive operators:
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
x∈E,
yn=QC[( –αn)(xn+en)], n≥,
zn= ( –βn)xn+βn[ayn+
N
i=ai(yn–rn,iBiyn)], n≥,
xn+=γnηf(xn) + (I–γnT)zn, n≥.
(D)
Let E,C,QC,f,k,η,T,J,γ,am(m= , , . . . ,N),{en},{αn},{βn},and{γn},{rn,i}satisfy the
same conditions as those in Theorem.Let Bi:C→E be anα-inversely strongly accretive
operator satisfying for∀r> and i= , , . . . ,N,
Bix–Biy,J
(I–rBi)x– (I–rBi)y
≥.
IfNi=B–i =∅,then{xn}generated by(D)converges strongly to a point p∈
N
i=B–i ,
which is the unique solution of the following variational inequality:for∀z∈Ni=B–
i ,
(T–ηf)p,J(p–z)
≤. (∗∗∗∗)
3 Connection with nonlinear capillarity equation
Remark In the next of this paper, we have four purposes: () give a new example to show that the assumption that ‘the set of zeros of the sum of anm-accretive operator and anα-inversely strongly monotone operator is nonempty’ is valid; that is, (A+B)–=
Remark In the following, assume NN+<p< +∞, ≤q,r< +∞ifp≥N, and ≤q,r≤
Np
N–p ifp<N, whereN≥. · pdenotes the norm inLp(). Let
p+
p = .
We shall examine the following capillarity equation, which is a special case in []:
⎧ ⎪ ⎨ ⎪ ⎩
–div[( +√|∇u|p +|∇u|p)|∇u|
p–∇u] +λ(|u|q–u+|u|r–u) +u(x) = , a.e. in, –ϑ, ( +√|∇u|p
+|∇u|p)|∇u|
p–∇u= , a.e. on, (E)
whereis a bounded conical domain of a Euclidean spaceRN with its boundary∈C (cf.[]).| · |denotes the Euclidean norm inRN,·,·the Euclidean inner-product andϑ
the exterior normal derivative of.λis a nonnegative constant.
Theorem (see []) The capillarity equation(E)has a unique solution u(x)∈Lp().
Lemma (see []) Define the mapping Bp,q,r:W,p()→(W,p())∗by
v,Bp,q,ru=
+ |∇u|
p
+|∇u|p
|∇u|p–∇u,∇v
dx
+λ
u(x)q–u(x)v(x)dx
+λ
u(x)r–u(x)v(x)dx,
for any u,v∈W,p().Then B
p,q,r is everywhere defined,strictly monotone,
hemi-contin-uous and coercive.
Lemma (see []) Define a mapping A:Lp()→Lp()
as follows:
D(A) =u∈Lp()|there exists an f∈Lp(),such that f ∈Bp,q,ru
.
For u∈D(A),let Au={f ∈Lp()|f ∈B p,q,ru}.
Then A:Lp()→Lp()
is m-accretive.
Lemma Define a mapping C:Lp()→Lp()by Cu=u(x),for∀u(x)∈Lp().
Then C is-inversely strongly accretive.
Proof LetJp:Lp()→Lp
() denote the normalized duality mapping. Then it is easy to check thatJpu=|u|p–sgnuu–p p,∀u∈Lp().
Thus for∀u(x),v(x)∈Lp(),Cu–Cv,J
p(u–v)=
|u–v|
pu–v–p
p dx=u–vp,
which implies thatCis -inversely strongly accretive.
This completes the proof.
Theorem u(x)∈Lp()is the unique solution of(E)if and only if u(x)∈(A+C)–.
Proof Ifu(x) is the solution of (E), then
–div
+ |∇u|
p
+|∇u|p
|∇u|p–∇u
Thus for∀ϕ∈C∞(), by using the property of generalized functions, we have
=
ϕ, –div
+ |∇u|
p
+|∇u|p
|∇u|p–∇u
+λ|u|q–u+|u|r–u+u(x)
= –div
+ |∇u|
p
+|∇u|p
|∇u|p–∇u ϕdx +
λ|u|q–u+|u|r–u+u(x)ϕdx
=
+ |∇u|
p
+|∇u|p
|∇u|p–∇u,∇ϕ
dx +
λ|u|q–u+|u|r–u+u(x)ϕdx
=ϕ,Bp,q,ru+Cu=ϕ,Au+Cu.
Thenu(x)∈(A+C)–.
On the other hand, ifu(x)∈(A+C)–, then for∀ϕ∈C∞ (),
=ϕ,Au+Cu=ϕ,Bp,q,ru+Cu
=
+ |∇u|
p
+|∇u|p
|∇u|p–∇u,∇ϕ
dx +λ
|u|q–u+|u|r–uϕdx+
u(x)ϕdx
=
ϕ, –div
+ |∇u|
p
+|∇u|p
|∇u|p–∇u
+λ|u|q–u+|u|r–u+u(x)
.
Then –div[( +√|∇u|p +|∇u|p)|∇u|
p–∇u] +λ(|u|q–u+|u|r–u) +u(x) = , a.e.x∈. By using the Green’s formula, we know that for anyv∈W,p(),
ϑ,
+ |∇u|
p
+|∇u|p
|∇u|p–∇u
v|d(x)
=
div
+ |∇u|
p
+|∇u|p
|∇u|p–∇u v dx +
+ |∇u|
p
+|∇u|p
|∇u|p–∇u,∇v dx =
λ|u|q–u+|u|r–u+u(x)dx+v,Bp,q,ru
–
λ|u|q–u+|u|r–udx
=v,Au+Cu= .
Thus –ϑ, ( +√|∇u|p +|∇u|p)|∇u|
Thenu(x)∈(A+C)– implies thatu(x) is the solution of (E).
This completes the proof.
Theorem Suppose A and C are the same as those in Lemmasand,respectively.
Let T:Lp()→Lp()be any strongly positive linear bounded operator with coefficientγ
and f :Lp()→Lp()be a contraction with coefficient k.Suppose the following conditions
are satisfied:
(i) <η<γk,and <a< ; (ii) {en}∞n=⊂Lp(),
∞
n=en< +∞;
(iii) {αn},{βn},and{γn}are three sequences in(, ).γn→, γγnn–→,βn→,αn→,
asn→ ∞.∞n=γn=∞,
∞
n=|αn+–αn|< +∞,
∞
n=|βn+–βn|< +∞,
∞
n=( –γnγ)αnβn< +∞;
(iv) {rn} ⊂(, )such that
∞
n=|rn+–rn|< +∞,and≥rn≥ε> forn≥.
If we construct the following iterative algorithm:
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
u(x)∈Lp(),
vn(x) = ( –αn)(un(x) +en(x)),
wn(x) = ( –βn)un(x) +βn[avn(x) + ( –a)JrAn(vn(x) –rnCvn(x))],
un+(x) =γnηf(un) + (I–γnT)wn(x), n≥,
(F)
then un(x)converges strongly to u(x)∈(A+C)–,which is the unique solution of the
capil-larity equation(E)and satisfies the following variational inequality:for∀z(x)∈(A+C)–,
(T–ηf)u(x),Jp
u(x) –z≤.
Remark From Theorem we can easily see the relationship among the solution of the capillarity equation, the solution of a variational inequality, and the zero of sum of an
m-accretive operator and anα-inversely strongly accretive operator.
Remark LetCbe the -inversely strong accretive operator defined in Lemma , then it is obvious thatCsatisfies
Cx–Cy,Jp
(I–rC)x– (I–rC)y≥, for ≥r> ,x,y∈Lp().
Thus the assumption imposed onBiin Theorem is valid.
Competing interests
The authors declare that there is no conflict of interests regarding the publication of this article.
Authors’ contributions
All authors contributed equally to this manuscript. All authors read and approved the final manuscript.
Acknowledgements
This paper is supported by the National Natural Science Foundation of China (No. 11071053), Natural Science Foundation of Hebei Province (No. A2014207010), Key Project of Science and Research of Hebei Educational Department
(ZH2012080) and Key Project of Science and Research of Hebei University of Economics and Business (2013KYZ01).
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